//给定一个单词数组 words 和一个长度 maxWidth ，重新排版单词，使其成为每行恰好有 maxWidth 个字符，且左右两端对齐的文本。 
//
// 你应该使用 “贪心算法” 来放置给定的单词；也就是说，尽可能多地往每行中放置单词。必要时可用空格 ' ' 填充，使得每行恰好有 maxWidth 个字符。
// 
//
// 要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配，则左侧放置的空格数要多于右侧的空格数。 
//
// 文本的最后一行应为左对齐，且单词之间不插入额外的空格。 
//
// 注意: 
//
// 
// 单词是指由非空格字符组成的字符序列。 
// 每个单词的长度大于 0，小于等于 maxWidth。 
// 输入单词数组 words 至少包含一个单词。 
// 
//
// 
//
// 示例 1: 
//
// 
//输入: words = ["This", "is", "an", "example", "of", "text", "justification."], 
//maxWidth = 16
//输出:
//[
//   "This    is    an",
//   "example  of text",
//   "justification.  "
//]
// 
//
// 示例 2: 
//
// 
//输入:words = ["What","must","be","acknowledgment","shall","be"], maxWidth = 16
//输出:
//[
//  "What   must   be",
//  "acknowledgment  ",
//  "shall be        "
//]
//解释: 注意最后一行的格式应为 "shall be    " 而不是 "shall     be",
//     因为最后一行应为左对齐，而不是左右两端对齐。       
//     第二行同样为左对齐，这是因为这行只包含一个单词。
// 
//
// 示例 3: 
//
// 
//输入:words = ["Science","is","what","we","understand","well","enough","to",
//"explain","to","a","computer.","Art","is","everything","else","we","do"]，maxWidth = 2
//0
//输出:
//[
//  "Science  is  what we",
//  "understand      well",
//  "enough to explain to",
//  "a  computer.  Art is",
//  "everything  else  we",
//  "do                  "
//]
// 
//
// 
//
// 提示: 
//
// 
// 1 <= words.length <= 300 
// 1 <= words[i].length <= 20 
// words[i] 由小写英文字母和符号组成 
// 1 <= maxWidth <= 100 
// words[i].length <= maxWidth 
// 
//
// Related Topics 数组 字符串 模拟 👍 461 👎 0

package com.cute.leetcode.editor.cn;

import java.util.*;

public class TextJustification {
    public static void main(String[] args) {
        Solution solution = new TextJustification().new Solution();
        String[] words = new String[]{"Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"};
        solution.fullJustify(words, 20);
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        /**
         * 很直接的写法，双指针去找最大的单词，之后遍历填充，最后一行进行左对齐处理
         */
        public List<String> fullJustify(String[] words, int maxWidth) {
            List<String> ans = new ArrayList<>();
            StringBuilder cur;
            int i = 0, j = 0, n = words.length, count = 0, blankCount = 0;
            while (i < n && j < n) {
                while (j < n) {
                    count += words[j].length();
                    if (count + (j - i) > maxWidth) {// j指向的单词不可取
                        cur = new StringBuilder();
                        count -= words[j].length();
                        blankCount = j - i - 1;
                        if (blankCount > 0) {
                            int mod = (maxWidth - count) % blankCount;
                            int div = (maxWidth - count) / blankCount;
                            for (int k = i; k < j; k++) {
                                cur.append(words[k]);
                                if (k != j - 1) {
                                    if (mod-- > 0) cur.append(" ");
                                    for (int l = 0; l < div; l++) {
                                        cur.append(" ");
                                    }
                                }
                            }
                        } else {
                            cur.append(words[i]);
                        }
                        while (cur.length() < maxWidth) cur.append(" ");
                        i = j;// 跳转到新单词
                        count = 0;
                        ans.add(cur.toString());
                    } else {
                        j++;
                    }
                    if (j == n) {// 最后一行左对齐处理
                        cur = new StringBuilder();
                        for (int k = i; k < j; k++)  cur.append(words[k]).append(" ");
                        if (cur.length() > maxWidth) cur.deleteCharAt(maxWidth);
                        while (cur.length() < maxWidth) cur.append(" ");
                        ans.add(cur.toString());
                    }
                }
            }
            return ans;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}